difficult sin x function

**pacman** · July 31st 2009, 07:10 PM

Solve for x: (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27

**Opalg** · August 1st 2009, 04:16 AM

Originally Posted by pacman

Solve for x: $(16/81)^{\sin^2 x} + (8/9)^{\cos^2 x} = 26/27$

Something wrong here? If you put $y = \sin^2x$ then $0\leqslant y\leqslant1$ . But the function $(16/81)^y + (8/9)^{1-y}$ is greater than 1 throughout the interval [0,1], so it can never be equal to 26/27.

**Krahl** · August 1st 2009, 04:44 PM

ok say instead of 26/27 it was 3/2.

how can you solve it for an exact soln? w/o using any approximating methods

**simplependulum** · August 2nd 2009, 01:38 AM

Originally Posted by Krahl

ok say instead of 26/27 it was 3/2.

how can you solve it for an exact soln? w/o using any approximating methods

If It is $( \frac{4}{9})^{\cos^2{x}}$ not

$( \frac{8}{9})^{\cos^2{x}}$

Sub $(\frac{4}{9})^{\sin^2{x}} = u$ then

$<br /> u^2 + (4/9) u^{-1} = M<br />$

$u^3 - M u + (4/9) = 0$

It becomes a cubic equation

after solving u $u = (\frac{4}{9})^{\sin^2{x}}$

$x = \arcsin{ \sqrt{ \frac{ \ln{u} }{ \ln(4/9) } } }$

**pacman** · August 10th 2009, 02:46 AM

i plot this one: y = (16/81)^(sin^2 x) + (8/9)^(cos^2 x) for x =0 to pi radian. In my graph it looks like this, any comments?

**pacman** · August 10th 2009, 02:50 AM

and this one, (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27; what does it mean?

**Failure** · August 10th 2009, 09:38 AM

Originally Posted by pacman

i plot this one: y = (16/81)^(sin^2 x) + (8/9)^(cos^2 x) for x =0 to pi radian. In my graph it looks like this, any comments?

My comment: it is not true that this is the plot of $y = (16/81)^{\sin^2 x} + (8/9)^{\cos^2 x}$ . Rather, this is the plot of $y = (16/81)^{sin^2 x} + (8/9)^{cos^2 x}-26/27$ .

Also: This function assumes a minimum value of $\tfrac{19}{18}\approx 0.235$ at $x=\tfrac{\pi}{2}+n\cdot\pi, n\in\mathbb{Z}$ . But it is always $>0$ .

**Failure** · August 10th 2009, 09:43 AM

Originally Posted by pacman

and this one, (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27; what does it mean?

It means that this time you have plotted $y=(16/81)^{sin^2 x} + (8/9)^{cos^2 x}$ for a larger range of x and, for some reason that I cannot fathom, seem to have surprised yourself with the result.

**pacman** · August 11th 2009, 07:12 AM

your observation is quite alright with me, thanks

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